Michael A. Nielsen

1
A geometric approach to lower bounds on quantum
circuit size
Michael A. Nielsen
School of Physical Sciences
The University of Queensland
To appear next week on quant-ph
MAQIS, UQ, Jan 28, 2005
2
A supposedly hard problem The traveling salesman
problem (TSP)
22 km
23 km
12 km
15 km
19 km
14 km
16 km
Goal Find the shortest tour through all the
cities.
It is widely believed that this problem requires
exponential time to solve.
Unfortunately, nobody can prove this.
In fact, we have embarassing little idea how to
prove this.
3
The P versus NP problem
P is (roughly) the class of problems soluble in
polynomial time on a classical computer.
NP is (roughly) the class of problems whose
solutions can be checked in polynomial time on a
classical computer.
TSP and factoring are examples of problems in NP,
but believed not to be in P.
Conjecture P ? NP
Smale (Fields medal, 1960) The most important
mathematical conjecture made in the second half
of the twentieth century.
Clay Millenium Prize problem
4
The P versus NP problem
It seems to be very difficult to prove that a
problem like TSP is hard to solve.
We expect TSP to take time 2n.
In fact, the best lower bounds on running time
essentially say You must read the input, i.e.,
they scale as the size of the problem.
5
Quantum computing 101 quantum circuits
0i
Built up out of qubits C2
0i
A single line a quantum wire represents a
single qubit.
0i
0i
Progress through time is from left to right.
0i
0i
0i
Fix a standard computational basis,0i, 1i, for
the qubit.
0i
Quantum computer is made up of n qubits C2
Initially, the qubits are all in some standard
state, say 0in.
Dynamics One- and two-qubit unitary gates.
Conclude by measuring some of the qubits in their
computational bases.
Knill Most unitaries require exponentially many
gates to approximate.
Church-Turing-Deutsch Principle (rough) If it
cant be done efficiently on a quantum computer,
it cant be done efficiently in nature.
6
Proving things are hard to do quantumly
Let U be a unitary operation on n qubits.
Find the minimal number of quantum gates needed
to implement U exactly.
Main point of the talk To sketch a geometric
program aimed at getting a handle on this minimal
number of gates.
7
Idea 1 Replace discrete structures by smooth
structures
f(x)
x
To minimize a function f(x) defined on the
integers weve got to solve the inequality f(x)
f(x-1),f(x1)
Suppose f(x) is defined on the reals.
Powerful principle Provided f is smooth, then
f(x) is stationary at any minimum.
Local minima can be found by solving a single
equality, f(x) 0.
Problem of finding minimal quantum circuits it
is tempting to think in terms of optimizing over
a discrete space of quantum circuits.
Better approach Replace circuits by
time-dependent Hamiltonians, and use the
principle of stationarity (via calculus of
variations) to identify local minima.
8
Idea 2 Find a metric space so the optimal
Hamiltonian control is a minimal geodesic of that
metric space
Motivated by general relativity (GR).
In GR a test particle follows geodesics of
spacetime.
Geodesics are paths which are (local) minima of
length.
Thus, the motion of a test particle minimizes a
global functional the length.
This is equivalent to following a local force
law known as the geodesic equation
acceleration
determined by local shape of spacetime
velocities
9
Idea 2 Find a metric space so the optimal
Hamiltonian control is a minimal geodesic of that
metric space
Motivated by general relativity (GR).
In GR a test particle follows geodesics of
spacetime.
Geodesics are paths which are (local) minima of
length.
Thus, the motion of a test particle minimizes a
global functional the length.
This is equivalent to following a local force
law known as the geodesic equation
Not special to GR In any Finsler metric space,
finding geodesics minimizing length is
equivalent to following a local force law.
The geodesic equation is a second order ODE, so
given an initial position and velocity,
subsequent motion is completely determined.
(Compare, e.g., with being given part of an
optimal circuit.)
10
Basic Picture
SU(2n)
We can think of U as being generated by an
n-qubit time-dependent Hamiltonian
H(t) ?? ??(t) ? V(0) I V(1) U
U
I
V(t)
The Hamiltonian H(t) generates small displacement
s near U(t).
We define a local metric F(V,H)
Length of the curve V lF(V) s01 dt
F(V(t),H(t)).
Distance between I and U dF(I,U) inf lF(V)
For any fixed V, F(V,) should be a norm F(V,H)
0 with equality iff H 0 F(V,?H) ?
F(V,H) F(V,H1H2) F(V,H1)F(V,H2)
Notation ? generalized Pauli tensor
product of Is, Xs, Ys, and Zs
11
Basic Picture
SU(2n)
We can think of U as being generated by an
n-qubit time-dependent Hamiltonian
H(t) ?? ??(t) ? V(0) I V(1) U
U
I
V(t)
The Hamiltonian H(t) generates small displacement
s near U(t).
We have to impose two additional conditions in
order to obtain a suitable geodesic equation.
1. F(V,H) smooth away from from H 0.
F is a Finsler metric on SU(2n)
2. Let y? be set of co-ordinates for H.
12
Basic Picture
SU(2n)
We can think of U as being generated by an
n-qubit time-dependent Hamiltonian
H(t) ?? ??(t) ? V(0) I V(1) U
U
I
V(t)
The Hamiltonian H(t) generates small displacement
s near U(t).
Key point Operationally, a Finsler metric (on
SU(2n)) corresponds to the type of metric for
which we can write down a geodesic equation to
study the minimal length paths.
13
Basic Picture
SU(2n)
We can think of U as being generated by an
n-qubit time-dependent Hamiltonian
H(t) ?? ??(t) ? V(0) I V(1) U
U
I
V(t)
The Hamiltonian H(t) generates small displacement
s near U(t).
We are minimizing lF(V) s01 dt F(V(t),H(t))
Introduce co-ordinates xj on SU(2n)
Let yj dxj/dt be the corresponding natural
velocity co-ordinates for the tangent space.
Euler-Lagrange equations from the calculus of
variations tell us that any minimizing curve must
satisfy
14
Basic Picture
SU(2n)
We can think of U as being generated by an
n-qubit time-dependent Hamiltonian
H(t) ?? ??(t) ? V(0) I V(1) U
U
I
V(t)
The Hamiltonian H(t) generates small displacement
s near U(t).
With some algebra and using the properties of F,
the Euler-Lagrange equations may be recast in the
form
Main point this is a second order ODE, and so
once the initial position and velocity are set we
have a unique solution.
Functions of local metric (both position and
velocity)
15
The relationship between geometry and circuits
SU(2n)
Universality We can build up U out of gates of
the form exp(-i? ?), where ? 1.
U
Let m(U) be the minimal number of gates of this
form necessary to build up U.
I
We can construct a control function ?(t) which
replicates this path.
Suppose F(V,?) 1 for all V and ?.
Key condition on F.
Then the distance along the induced path is at
most ?1?2?m(U) m(U)
It follows that dF(I,U) m(U)
16
Example
SU(2n)
H ?? ?? ?
U
I
V(t)
This is an example of a Riemannian metric on
SU(2n) a special type of Finsler
metric, arising from a quadratic form.
We have F2(V,?) 1, so dF2(I,U) m(U).
Unfortunately, dF2(I,U) ?2, and so the best
lower bound on m(U) we can hope for is constant.
Proof Choose H so that exp(-iH) U, and all
eigenvalues of H are between -? and ?.
Then the curve U(t) exp(-iHt) has length at
most tr(H2)/2n ?2.
17
Better examples
Penalty function
SU(2n)
H ?? ?? ?
U
I
V(t)
Both satisfy F(V,?) 1, so dF(I,U) m(U).
F1 is not Finsler, and so needs to be
appoximated by a Finsler metric.
Idea of approximation
For small ?, this becomes a good approximation to
F1.
The idea is to choose F1? so that unit vectors
H satisfy g?(H) 1.
F1? defined in this way is a Finsler metric.
18
Summary
SU(2n)
Weve constructed a metric F1(V,H) on SU(2n)
which gives lower bounds on circuit
size, dF1(I,U) m(U)
U
I
V(t)
We write F1(H) F1?(V,H)
F1 can be approximated by a family of Finsler
metrics, F1?
Minimal length curves for F1? must be solutions
of the geodesic equation.
The geodesic equation is not trivial to compute,
nor to solve!
Everything I say about F1 for the remainder of
the talk is also true of Fq, provided q is chosen
appropriately.
19
Finding geodesics on the sphere
Lets choose a starting point.
And a starting velocity.
The geodesic equation tells us there is a unique
geodesic motion determined by these choices.
Pick the pole going through the the starting
point.
Consider the plane spanned by the pole and the
initial velocity.
If we reflect the sphere through that plane,
lengths are not changed.
So the reflected geodesic is still a geodesic.
The reflected geodesic has same starting point
and velocity ) its the same geodesic!
Must be a great circle.
20
Finding geodesics on SU(2n)
SU(2n)
Choose a starting point.
And a starting Hamiltonian, H, diagonal in the
computational basis.
The geodesic equation tells us there is a unique
geodesic motion determined by these choices.
I
Let ? be a Pauli matrix.
The map U ! ? U ? is an isometry of F1?.
If V(t) is the geodesic, then ? V(t) ? is also a
geodesic.
If ? is a product of Is and Zs, then ? V(t) ? has
the same starting point and starting Hamiltonian.
V(t) is diagonal in the computational basis.
A simple computation based on the geodesic
equation shows that V(t) exp(-i Ht).
21
Pauli geodesics on SU(2n)
SU(2n)
H diagonal in the computational basis.
U
Geodesic is V(t) exp(-i Ht).
I
Given any U diagonal in computational basis, we
can find a Pauli geodesic through U H i ln(U)
) V(1) U
Given such a Pauli geodesic, the length between I
and U is lF1 F1(H)
For a given U, there are typically many Pauli
geodesics from I through U.
exp(-iHt), where H H-J, and J is diagonal in
the computational basis, with entries which are
integer multiples of 2?.
Q1 Which Pauli geodesic through a given U is the
shortest?
Q2 Is the shortest Pauli geodesic also the
shortest geodesic?
22
Freedom in the Pauli geodesics
SU(2n)
U diagonal in computational basis
U
Fix H such that U exp(-iH)

• Find J such that
• J diagonal in comp basis, with entries2 ?
  integers.
• J minimizes F1(H-J)

I
The set of allowed J forms a lattice that is,
it is closed under integer linear combinations
n1 J1 n2 J2
The problem of minimizing F(H-J) is thus the
problem of finding the closest lattice element J
to H, under the norm F1.
Main point Finding the minimum length Pauli
geodesic is an instance of solving a problem well
known in computer science, the closest vector to
a lattice problem.
23
Consequences
A volume argument shows that all but a tiny
fraction of U have exponentially long minimal
Pauli geodesics.
Q Can we explicitly construct a family of
unitaries U that haveexponentially long minimal
length Pauli geodesics?
When is the shortest Pauli geodesic from I to
U the shortest among all geodesics?
A In general, I dont know. But we do have
Proposition Suppose U is diagonal in the
computational basis. If the shortest geodesic
from I to U is unique then it must be a Pauli
geodesic.
Proof Similar argument to that used earlier if
it is unique, then it must commute with Paulis
which are products of I and Z, and so must be
diagonal in computational basis ) Pauli geodesic.
24
(No Transcript)